
after applying [1]}. Note that
\(& \frac{1}{\bar{b}}\left(\bar{{\mathbf {R}}}_{g}{\mathbf {\Sigma }}_{g}\bar{{\mathbf {R}}}_{g}{\mathbf {\Sigma }}_{g}\right)=\frac{1}{\bar{b}}\left(\bar{{\mathbf {R}}}_{g}^{-1/2}\bar{{\mathbf {R}}}_{g}\bar{{\mathbf {R}}}_{g}^{-1/2}{\mathbf {\Sigma }}_{g}^{2}\right)\nonumber \\&={\mathbf {d}}{}{z}\frac{1}{\bar{b}}\left(\bar{{\mathbf {R}}}_{g}\left(\frac{1}{\bar{b}}{\mathbf {B}}_{g}^{{\scriptscriptstyle \mathsf {H}}}\hat{\bar{{\mathbf {H}}}}_{g}\hat{\bar{{\mathbf {H}}}}_{g}{\mathbf {B}}_{g}+\alpha {\mathbf {\mathrm {I}}}_{\bar{b}}-z\bar{{\mathbf {R}}}_{g}\right)^{-1}\right)_{z=0}\nonumber \\&\asymp \frac{1}{\bar{b}}\left(\bar{{\mathbf {R}}}_{g}\bar{{\mathbf {T}}}_{g}^{\prime }(0)\right),\) 
